Медіани АК-12см та ВМ-15см трикутника АВС. Знайдіть відрізки АО,ОК, ВО, ОМ, де О-точка перетину медіан трикутника АВС обьясните как вы решили задачу

ответ: сторона квадрата равна 22.

Пошаговое объяснение:

Пусть сторона квадрата равна х. Если одну из сторон квадрата увеличить на 5, а соседнюю уменьшить на 3, то получим прямоугольник со сторонами х+5 и х-3.

Площадь квадрата равна: S=х²

Площадь прямоугольника равна: (х+5)(х-3) и на 29 больше площади квадрата.

Составим и решим уравнение:

(х+5)(х-3)-х²=29

х²+5х-3х-15-х²=29

2х-15=29

2х=29+15

2х=44

х=44:2

х=22  - сторона квадрата.

Проверим:

Площадь квадрата: 22²=484

Площадь прямоугольника: (22+5)(22-3)=27*19=513

513-484=29

Solve for x over the real numbers:
x^2-3 x+2 = 2 sqrt(x) (x-1)
Subtract 2 sqrt(x) (x-1) from both sides:
2-2 sqrt(x) (x-1)-3 x+x^2 = 0
2-2 (x-1) sqrt(x)-3 x+x^2 = 2+2 sqrt(x)-3 x-2 x^(3/2)+x^2:
2+2 sqrt(x)-3 x-2 x^(3/2)+x^2 = 0
Simplify and substitute y = sqrt(x): 2+2 sqrt(x)-3 x-2 x^(3/2)+x^2 = 2+2 sqrt(x)-3 (sqrt(x))^2-2 (sqrt(x))^3+(sqrt(x))^4 = y^4-2 y^3-3 y^2+2 y+2 = 0:
y^4-2 y^3-3 y^2+2 y+2 = 0
The left hand side factors into a product with three terms:
(y-1) (y+1) (y^2-2 y-2) = 0
Split into three equations:
y-1 = 0 or y+1 = 0 or y^2-2 y-2 = 0
Add 1 to both sides:
y = 1 or y+1 = 0 or y^2-2 y-2 = 0
Substitute back for y = sqrt(x):
sqrt(x) = 1 or y+1 = 0 or y^2-2 y-2 = 0
Raise both sides to the power of two:
x = 1 or y+1 = 0 or y^2-2 y-2 = 0
Subtract 1 from both sides:
x = 1 or y = -1 or y^2-2 y-2 = 0
Substitute back for y = sqrt(x):
x = 1 or sqrt(x) = -1 or y^2-2 y-2 = 0
Raise both sides to the power of two:
x = 1 or x = 1 or y^2-2 y-2 = 0
Add 2 to both sides:
x = 1 or x = 1 or y^2-2 y = 2
Add 1 to both sides:
x = 1 or x = 1 or y^2-2 y+1 = 3
Write the left hand side as a square:
x = 1 or x = 1 or (y-1)^2 = 3
Take the square root of both sides:
x = 1 or x = 1 or y-1 = sqrt(3) or y-1 = -sqrt(3)
Add 1 to both sides:
x = 1 or x = 1 or y = 1+sqrt(3) or y-1 = -sqrt(3)
Substitute back for y = sqrt(x):
x = 1 or x = 1 or sqrt(x) = 1+sqrt(3) or y-1 = -sqrt(3)
Raise both sides to the power of two:
x = 1 or x = 1 or x = (1+sqrt(3))^2 or y-1 = -sqrt(3)
Add 1 to both sides:
x = 1 or x = 1 or x = (1+sqrt(3))^2 or y = 1-sqrt(3)
Substitute back for y = sqrt(x):
x = 1 or x = 1 or x = (1+sqrt(3))^2 or sqrt(x) = 1-sqrt(3)
Raise both sides to the power of two:
x = 1 or x = 1 or x = (1+sqrt(3))^2 or x = (1-sqrt(3))^2
x^2-3 x+2 => 2-3 1+1^2 = 0
2 (x-1) sqrt(x) => 2 (1-1) sqrt(1) = 0:
So this solution is correct
x^2-3 x+2 => 2-3 (1-sqrt(3))^2+((1-sqrt(3))^2)^2 = 18-10 sqrt(3) ~~ 0.679492
2 (x-1) sqrt(x) => 2 ((1-sqrt(3))^2-1) sqrt((1-sqrt(3))^2) = 2 (5 sqrt(3)-9) ~~ -0.679492:
So this solution is incorrect
x^2-3 x+2 => 2-3 (1+sqrt(3))^2+((1+sqrt(3))^2)^2 = 2 (9+5 sqrt(3)) ~~ 35.3205
2 (x-1) sqrt(x) => 2 ((1+sqrt(3))^2-1) sqrt((1+sqrt(3))^2) = 2 (9+5 sqrt(3)) ~~ 35.3205:
So this solution is correct
The solutions are:
Answer: |
| x = 1 or x = 1 or x = (1+sqrt(3))^2